Question

Consider the hydrogen atom to be a proton embedded in a cavity of radius $a_0$ (Bohr’s radius) whose charge is neutralized by the addition of an electron to the cavity in vacuum, infinitely slowly. If the magnitude of the average kinetic energy is half the magnitude of the average energy, the average potential energy:

• A. $\frac{-2e^2}{\pi {ϵ}_o{a}_o}$
• B. $\frac{-e^2}{3\pi {ϵ}_o{a}_o}$
• C. $\frac{-e^2}{2\pi {ϵ}_o{a}_o}$
• D. $\frac{-2e^2}{\pi {ϵ}_o{a}_o}$

Answer 1

The correct option is C $\frac{-e^2}{2\pi {ϵ}_o{a}_o}$

The work done in the given neutralisation process is:
$W=-{\int }_{a_o}^{\infty }Fdr$
where, $F=\frac{e^2}{4\pi {ϵ}_o{r}^2}$
and $r$ = radius of the atom
$W=\frac{e^2}{4\pi {ϵ}_o}[\frac{1}{r}{]}_{a_o}^{\infty }$
= $\frac{-e^2}{4\pi {ϵ}_o{a}_o}$ = Total energy $\left(E\right)$
Now, if "$V$" is magnitude of potential energy, then according to given information, kinetic energy (Ek) is V/2. Therefore,
$E=-V+\frac{V}{2}$
= $\frac{-V}{2}$
$V=-2E=\frac{e^2}{2\pi {ϵ}_o{a}_o}$
Hence the correct option is $c$